1) The document analyzes rigidly fixed portal frames and develops equations to calculate internal stresses considering axial deformation.
2) It uses the flexibility method and principle of virtual work to obtain the flexibility matrix and circumvent directly inverting the stiffness matrix.
3) The flexibility matrix is used to determine redundant forces and the internal stresses are calculated through equilibrium equations and superposition of stresses. Considering axial deformation provides a more accurate analysis of portal frames.
Discrete Model of Two Predators competing for One Prey
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Contribution of Axial Deformation in the Analysis of Rigidly Fixed Portal Frames
1. International Journal of Modern Engineering Research (IJMER)
www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645
Contribution of Axial Deformation in the Analysis of Rigidly
Fixed Portal Frames
Okonkwo V. O.1, Aginam C. H.2, and Chidolue C. A.3
123
Department of Civil Engineering Nnamdi Azikiwe University, Awka Anambra State-Nigeria
ABSTRACT: In this work the stiffness equations for
evaluating the internal stress of rigidly fixed portal frames
by the displacement method were generated. But obtaining
the equations for the internal stresses required non-scalar
or parametric inversion of the structure stiffness matrix. To
circumvent this problem, the flexibility method was used
taking advantage of the symmetrical nature of the portal
frame and the method of virtual work. These were used to
obtain the internal stresses on rigidly fixed portal frames
for different cases of external loads when axial deformation
is considered. A dimensionless constant s was used to
capture the effect of axial deformation in the equations.
When it is set to zero, the effect of axial deformation is
ignored and the equations become the same as what can be
obtained in any structural engineering textbook. These
equations were used to investigate the contribution of axial
deformation to the calculated internal stresses and how
they vary with the ratio of the flexural rigidity of the beam
and columns and the height to length ratio of the loaded
portal frames.
Figure 1: A simple portal frame showing its dimensions and
Keywords: Axial deformation, flexural rigidity, flexibility
the 6 degrees of freedom
method, Portal frames, stiffness matrix,
The analysis of portal frames by the stiffness method
I. INTRODUCTION requires the determination of the structureβs degrees of
Structural frames are primarily responsible for
freedom and the development of the structureβs stiffness
strength and rigidity of buildings. For simpler single storey
matrix. For the structure shown in Figure 1a, the degrees of
structures like warehouses, garages etc portal frames are
freedom are as shown in Figure 1b. I1 and A1 are
usually adequate. It is estimated that about 50% of the hot- respectively the second moment of inertia and cross-
rolled constructional steel used in the UK is fabricated into sectional area of the columns while I2 and A2 are the second
single-storey buildings (Graham and Alan, 2007). This moment of inertia and cross-sectional area of the beam
shows the increasing importance of this fundamental
respectively. The stiffness coefficients for the various
structural assemblage. The analysis of portal frames are
degrees of freedom considering shear deformation can be
usually done with predetermined equations obtained from obtained from equations developed in Ghali et al (1985)
structural engineering textbooks or design manuals. The
and Okonkwo (2012) and are presented below
equations in these texts were derived with an underlying
assumption that deformation of structures due to axial The structureβs stiffness matrix can be written as:
forces is negligible giving rise to the need to undertake this
study. The twenty first century has seen an astronomical use π11 π12 π13 π14 π15 π16
of computers in the analysis of structures (Samuelsson and π21 π22 π23 π24 π25 π26
Zienkiewi, 2006) but this has not completely eliminated the π π32 π33 π34 π35 π36
use of manual calculations form simple structures and for πΎ = 31 . (1)
π41 π42 π43 π44 π45 π46
easy cross-checking of computer output (Hibbeler, 2006). π51 π52 π53 π54 π55 π56
Hence the need for the development of equations that π61 π62 π63 π64 π65 π66
capture the contribution of axial deformation in portal
frames for different loading conditions. Where kij is the force in coordinate (degree of freedom) i
when there is a unit displacement in coordinate (degree of
II. DEVELOPMENT OF THE MODEL freedom) j. They are as follows:
πΈπ΄1
π11 = β
π12 = 0
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2. International Journal of Modern Engineering Research (IJMER)
www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645
6πΈπΌ2 π1
π13 = π2
π2
12πΈπΌ1 π3
π14 = β β3
{π·} = . . . . . (5)
π4
π5
π15 = 0
π6
6πΈπΌ2
π16 = π2 Where di is the displacement in coordinate i.
12πΈπΌ1 πΈπ΄2 πΉ1
π22 = β3
+ π
πΉ2
6πΈπΌ1 πΉ3
π23 = {πΉ} = . . . . . . (6)
β2 πΉ4
πΉ5
π24 = 0
πΉ6
πΈπ΄2
π25 = β π
. . (2) Where Fi is the external load with a direction coinciding
with the coordinate i (McGuire et al, 2000).
π26 = 0
By making {D} the subject of the formula in equation (3)
4πΈπΌ2 4πΈπΌ1
π33 = +
π β β1
π· = πΎ {πΉ β πΉπ } . . . . (7)
6πΈπΌ2
π34 = β π2 Once {D} is obtained the internal stresses in the frame can
be easily obtained by writing the structureβs compatibility
π35 = 0 equations given as
2πΈπΌ2
π36 = π
π = π π + π1 π1 + π2 π2 + π3 π3 . . (8)
12πΈπΌ2 πΈπ΄1 Where M is the internal stress (bending moment) at any
π44 = +
π3 β point on the frame, Mr is the internal stress at the point
under consideration in the restrained structure while Mi is
π45 = 0 the internal stress at the point when there is a unit
6πΈπΌ2 displacement in coordinate i.
π46 = β To solve equation (8) there is need to obtain {D}. {D} can
π2
be obtained from the inversion of [K] in equation (7).
12πΈπΌ1 πΈπ΄2
π55 = + Finding the inverse of K parametrically (i.e. without
β3 π
substituting the numerical values of E, h, l etc) is a difficult
From Maxwellβs Reciprocal theorem and Bettiβs Law k ij = task. This problem is circumvented by using the flexibility
kji ( Leet and Uang, 2002). method to solve the same problem, taking advantage of the
symmetrical nature of the structure and the principle of
When there are external loads on the structure on the virtual work.
structure there is need to calculate the forces in the
restrained structure Fo as a result of the external load. III. APPLICATION OF THE FLEXIBILITY
MODEL
The structureβs equilibrium equations are then written as The basic system or primary structure for the structure in
Figure 1a is given in Figure 2. The removed redundant
{πΉ} = {πΉπ } + πΎ {π·} . . . . (3) force is depicted with X1.
π10
π20
π30
{πΉπ } = . . . . . (4)
π40
π50
π60
Where kio is the force due to external load in coordinate i
when the other degrees of freedom are restrained.
Figure 2: The Basic System showing the removed
redundant forces
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3. International Journal of Modern Engineering Research (IJMER)
www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645
The flexibility matrix of the structure can be determined 1 12πΈπΌ1 πΌ2 π΄1
= . . . (15a)
π 11 πΌ1 π΄1 π 3 +6πΌ2 π΄1 π 2 β +24πΌ1 πΌ2 β
using the principle of virtual work.
By applying the unit load theorem the deflection in beams
or frames can be determined for the combined action of the
internal stresses, bending moment and axial forces with π 33 3πΈπΌ1 π΄2 (ππΌ1 +2βπΌ2 )
π π π π 2
π 22 π 33 βπ 23
= 2π΄ 3 2 2 4 . (15b)
π·= ππ + ππ . . . (9) 2 β ππΌ1 +3πΌ1 π +π΄2 β πΌ2 +6βππΌ1 πΌ2
πΈπΌ πΈπ΄
π 32 β3πΈπΌ1 πΌ2 πΊπ΄2 β 2
Where π and π are the virtual internal stresses while M 2
π 22 π 33 βπ 23
= 2π΄ 3 2 2 4 (15c)
2 β ππΌ1 +3πΌ1 π +π΄2 β πΌ2 +6βππΌ1 πΌ2
and N are the real/actual internal stresses.
π 22 πΈπΌ1 πΌ2 2π΄2 β 3 +3πΌ1 π
E is the modulus of elasticity of the structural material 2 = 3 ππΌ +3πΌ 2 π 2 +π΄ β 4 πΌ +6βππΌ πΌ . (15d)
π 22 π 33 βπ 23 2π΄2 β 1 1 2 2 1 2
A is the cross-sectional area of the element (McGuire et al, Equation (12) is evaluated to get the redundant forces and
2000; Nash, 1998) these are substituted into the structures force equilibrium
(superposition) equations to obtain the internal stresses at
If dij is the deformation in the direction of i due to a unit any point.
load at j then by evaluating equation (9) the following are
obtained: π = π π + π1 π1 + π2 π2 + π3 π3 . . (16)
πΌ1 πΊπ΄1 π 3 +6πΌ2 π΄1 π 2 β +12πΌ1 πΌ2 β Where M is the required stress at a point, Mo is the stress at
π11 = . . (10a)
12πΈπΌ1 πΌ2 π΄1 that point for the reduced structure, Mi is stress at that point
when the only the redundant force Xi =1 acts on the reduced
π12 = 0 . . . . . (10b) structure.
For the loaded portal frame of Figure 3, the deformations of
π13 = 0 . . . . . . (10c)
the reduced structure due to external loads are
2π΄2 β 3 +3πΌ1 π
π22 = 3πΈπΌ1 π΄2
. . . (10d) π10 = 0 . . . . .
. (17a)
ββ 2
π23 = πΈπΌ1
. . . (10e) π€ β 2 π2
π20 = . . . . .
8πΈπΌ1
ππΌ1 +2β πΌ2 . (17b)
π33 = . . . . (10f)
πΈπΌ1 πΌ2
βπ€ π 2 (ππΌ1 +6βπΌ2 )
From Maxwellβs Reciprocal theorem and Bettiβs Law dij = π30 = . . (17c)
24πΈπΌ1 πΌ2
dji.
By substituting the values of equations (17a) β (17c) into
The structureβs compatibility equations can be written thus equations (12)
π11 π11 π11 π1 π10 π1 = 0 . . . . . (18a)
π21 π21 π21 π2 + π20 = 0 . (11a)
π31 π31 π31 π3 π30 βπ΄2 πΌ1 π€β 2 π 3
π2 = 4 2
2π΄2 β 3 ππΌ1 +3πΌ1 π 2 +π΄2 β 4 πΌ2 +6βππΌ1 πΌ2
. (18b)
i.e. ππΉ + π π = 0 . . . (11b) 2
π€ π 2 2π΄2 β 3 ππΌ1 +3πΌ1 π 2 +3π΄2 β 4 πΌ2 +18βππΌ1 πΌ2
π3 = 2
24 2π΄2 β 3 ππΌ1 +3πΌ1 π 2 +π΄2 β 4 πΌ2 +6βππΌ1 πΌ2
. (18c)
Where F is the vector of redundant forces X1, X2, X3 and do
is the vector deformation d10, d20, d30 due to external load
Evaluating equation (16) for different points on the
on the basic system (Jenkins, 1990).
structure using the force factors obtained in equations (18a)
πΉ = π β1 (βπ π ) . . . . (12) β (18c)
2π€π 3 πΌ1 β 3 π΄2 β3ππΌ1
π π΄ = 12 2
2π΄2 β 3 ππΌ1 +3πΌ1 π 2 +π΄2 β 4 πΌ2 +6βππΌ1 πΌ2
. . (19)
π΄ππ π
π β1 = πππ‘ π
. . (13)
β2π€π 3 πΌ1 2β 3 π΄2 +3ππΌ1
(Stroud, 1995) π π΅ = 24 2 . . (20)
2π΄2 β 3 ππΌ1 +3πΌ1 π 2 +π΄2 β 4 πΌ2 +6βππΌ1 πΌ2
1
π 11
0 0
π 33 βπ 32
π β1 = 0 2
π 22 π 33 βπ 23 2
π 22 π 33 βπ 23 . (14) β2π€ π 3 πΌ1 2β 3 π΄2 +3ππΌ1
ππΆ = 2 . (21)
βπ 23 π 22 24 2π΄2 β 3 ππΌ1 +3πΌ1 π 2 +π΄2 β 4 πΌ2 +6βππΌ1 πΌ2
0 2 2
π 22 π 33 βπ 23 π 22 π 33 βπ 23
βπ€ π 3 πΌ1 8πΌ2 β 3 +π 2 π 3 πΌ1
ππΆ =
From equations 10a β 10f 12 4β 3 πΌ2 2ππΌ1 +β πΌ2 +π 2 π 3 πΌ1 ππΌ1 +2β πΌ2
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6. International Journal of Modern Engineering Research (IJMER)
www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645
ignored π 2 = 0 . If axial deformation is ignored in the (about 40% drop in calculated bending moment value)
whole structure, π 1 = π 2 = 0. while at β πΏ > 0, βπ π΄ dropped in magnitude exponentially
The internal stress equations enable an easy investigation to values below 1.
into the contribution of axial deformation to the internal
stresses of statically loaded frames for different kinds of V. CONCLUSION
external loads. The flexibility method was used to simplify the analysis
For frame 1 (figure 3), the moment at A, MA is given by and a summary of the results are presented in table 1. The
equation (19). The contribution of axial deformation in the equations in table 1 would enable an easy evaluation of the
column βπ π΄ is given by internal stresses in loaded rigidly fixed portal frames
considering the effect of axial deformation.
βπ π΄ = π π΄(ππππ πππ’ππ‘πππ 19) β π π΄(ππππ πππ’ππ‘πππ 19 π€βππ π 2 =0) From a detailed analysis of frame 1 (Figure 3), it was
observed that the contribution of axial deformation is
π€ π 3 πΌ1 4β 3 πΌ2 +π 2 π 3 πΌ1 1 generally very small and can be neglected for reasonable
= 3 πΌ 2ππΌ +β πΌ +π π 3 πΌ ππΌ +2β πΌ β .
12 4β 2 1 1 2 1 1 2 2ππΌ1 +β πΌ2 values of πΌ1 πΌ2 . However, its contribution skyrockets at
. . (56) very low values of β π i.e. as β π => 0. This depicts the
case of an encased single span beam and a complete
Equation (56) gives the contribution of axial deformation to
departure from portal frames under study. This analysis can
MA as a function of h, l, I1, I2 and s
be extended to the other loaded frames (Figures 4 β 8) using
By considering the case of a portal frame of length l = 5m,
the equations in Table 1. These would enable the
h = 4m. Equation 56 was evaluated to show how the
determination of safe conditions for ignoring axial
contribution of axial deformation varied with πΌ1 πΌ2 . The
deformation under different kinds of loading.
result is shown in Table 2. When plotted on a uniform scale
(Figure 9) the relationship between βπ π΅ and πΌ1 πΌ2 is seen REFERENCES
to be linear. This was further justified by a linear regression
analysis of the results in Table 2 which was fitted into the [1] Ghali A, Neville A. M. (1996) Structural Analysis: A
πΌ Unified Classical and Matrix Approach (3rd Edition)
model βπ π΄ = π1 πΌ1 + π2 to obtain π1 = β0.004815 and Chapman & Hall London
2
π2 = 0.0001109 and the fitness parameters sum square of [2] Graham R, Alan P.(2007). Single Storey Buildings:
errors (SSE), coefficient of multiple determination (R2 ) and Steel Designerβs Manual Sixth Edition, Blackwell
the root mean squared error (RMSE) gave 1.099 x 10 -7, Science Ltd, United Kingdom
0.999952 and 0.0001105 respectively. From Table 2 when [3] Hibbeler, R. C.(2006). Structural Analysis. Sixth
πΌ1 πΌ2 = 0 , βπ π΄ β 0 and when πΌ1 πΌ2 = 10; βπ π΄ = Edition, Pearson Prentice Hall, New Jersey
β0.0479 which represent only a 5% reduction in the [4] Leet, K. M., Uang, C.,(2002). Fundamental of
calculated bending moment. Structural Analysis. McGraw-Hill ,New York
In like manner by evaluating the axial contribution in the
beam, βπ π΅ for varying πΌ1 πΌ2 of the portal frame, Table 3
[5] McGuire, W., Gallagher R. H., Ziemian, R.
was produced. This was plotted on a uniform scale in D.(2000). Matrix Structural Analysis,Second
Figure 10. From Figure 10 it would be observed that there Edition, John Wiley & Sons, Inc. New York
is also a linear relationship between βπ π΅ and πΌ1 πΌ2 . When [6] Nash, W.,(1998). Schaumβs Outline of Theory and
πΌ Problems of Strength of Materials. Fourth Edition,
fitted into the model βπ π΅ = π3 πΌ1 + π4 it gave π3 =
2 McGraw-Hill Companies, New York
β0.006502 and π4 = β0.0003969 for the fitness [7] Okonkwo V. O (2012). Computer-aided Analysis of
parameters sum square of errors (SSE), coefficient of Multi-storey Steel Frames. M. Eng. Thesis,
multiple determination (R2 ) and the root mean squared Nnamdi Azikiwe University, Awka, Nigeria.
error (RMSE) of 6.51 x 10-7, 0.9999 and 0.000269
[8] Reynolds, C. E.,and Steedman J. C. (2001).
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By pegging πΌ1 πΌ2 to a constant value of 0.296 and the Edition) E&FN Spon, Taylor & Francis Group,
variation of βπ π΄ with respect to β πΏ investigated, Table 4 London
was generated. A detailed plot of Table 4 was presented in
Figure 11. From Table 4 when β πΏ = 0; βπ π΄ = β3.125
[9] Samuelsson A., and Zienkiewicz O. C.(2006),
Review: History of the Stiffness Method.
(about 30% drop in calculated bending moment value)
International Journal for Numerical Methods in
while at β πΏ > 0, βπ π΄ dropped in magnitude exponentially
Engineering. Vol. 67: 149 β 157
to values below 1. In like manner, when the variation of
βπ π΅ with respect to β πΏ was investigated, Table 5 was [10] Stroud K. A.,(1995). Engineering Mathematics.
generated. A detailed plot of Table 5 was presented in Fourth Edition, Macmillan Press Ltd, London.
Figure 12. From Table 5 when β πΏ = 0; βπ π΄ = β4.1667
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7. International Journal of Modern Engineering Research (IJMER)
www.ijmer.com Vol.2, Issue.6, Nov-Dec. 2012 pp-4244-4255 ISSN: 2249-6645
Table 1: Internal stresses for a loaded rigid frame
S/No LOADED FRAME REMARKS
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